2M +6HCl–>2MCl+3H2 Delta H= -638.0 Kj
HClg—>HCl aq= Delta H -74.8 Kj
H2+Cl2–>2HCl Delta H -1845.0 Kj
MCl3s–> MCl3 aq Delta H -249.0 Kj
Use the information above to determine the enthalpy of the following reaction.
2M +3Cl2–>2MCl3 Delta H =???
To determine the enthalpy of the reaction 2M + 3Cl2 → 2MCl3, you can use the concept of Hess's Law. Hess's Law states that the enthalpy change of a reaction is independent of the route taken.
First, we need to rewrite the given equations in a way that allows us to cancel out some of the compounds:
1. Multiply the second equation (HClg → HCl aq) by 2 to match the number of moles of HCl in the first equation.
2HClg → 2HCl aq: ΔH = -149.6 kJ
2. Multiply the third equation (H2 + Cl2 → 2HCl) by 2 and reverse its direction to match the number of moles of HCl in the first equation.
2HCl → H2 + Cl2: ΔH = +3690 kJ
3. Multiply the fourth equation (MCl3s → MCl3 aq) by 2 and reverse its direction to match the number of moles of MCl3 in the desired equation.
2MCl3 → 2MCl3 aq: ΔH = +498 kJ
Now, we can sum up the modified equations:
2M + 2HClg → 2MCl aq: ΔH = -299.2 kJ (twice the second equation)
2MCl aq + 2H2 + 2Cl2 → 2MCl3: ΔH = -7380 kJ (twice the sum of the third and fourth equations)
Adding these two new equations gives us the desired equation:
2M + 3Cl2 → 2MCl3: ΔH = -299.2 kJ + (-7380 kJ) = -7679.2 kJ
Therefore, the enthalpy change of the reaction 2M + 3Cl2 → 2MCl3 is -7679.2 kJ.